题目内容
(本小题满分12分)已知数列{an}的前n项和为Sn,且an是Sn与2的等差中项,数列{bn}中,b1=1,点P(bn,bn+1)在直线上。
(1)求a1和a2的值;
(2)求数列{an},{bn}的通项an和bn;
(3)设cn=an·bn,求数列{cn}的前n项和Tn.
(1)求a1和a2的值;
(2)求数列{an},{bn}的通项an和bn;
(3)设cn=an·bn,求数列{cn}的前n项和Tn.
解:(1)∵an是Sn与2的等差中项 ∴Sn=2an-2 ∴a1=S1=2a1-2,
解得a1="2" a1+a2=S2=2a2-2,解得a2="4"
(2)∵Sn=2an-2,Sn-1=2an-1-2,又Sn—Sn-1=an, ∴an=2an-2an-1,
又an≠0, ∴,即数列{an}是等比数列
∵a1=2,∴an=2n
∵点P(bn,bn+1)在直线x-y+2=0上,∴bn-bn+1+2=0,
∴bn+1-bn=2,即数列{bn}是等差数列,又b1=1,∴bn=2n-1,
(3)∵cn=(2n-1)2n ∴Tn=a1b1+ a2b2+····anbn=1×2+3×22+5×23+····+(2n-1)2n,
∴2Tn=1×22+3×23+····+(2n-3)2n+(2n-1)2n+1
则 -Tn=1×2+(2×22+2×23+···+2×2n)-(2n-1)2n+1,
即:-Tn=1×2+(23+24+····+2n+1)-(2n-1)2n+1,
∴Tn=(2n-3)2n+1+6
解得a1="2" a1+a2=S2=2a2-2,解得a2="4"
(2)∵Sn=2an-2,Sn-1=2an-1-2,又Sn—Sn-1=an, ∴an=2an-2an-1,
又an≠0, ∴,即数列{an}是等比数列
∵a1=2,∴an=2n
∵点P(bn,bn+1)在直线x-y+2=0上,∴bn-bn+1+2=0,
∴bn+1-bn=2,即数列{bn}是等差数列,又b1=1,∴bn=2n-1,
(3)∵cn=(2n-1)2n ∴Tn=a1b1+ a2b2+····anbn=1×2+3×22+5×23+····+(2n-1)2n,
∴2Tn=1×22+3×23+····+(2n-3)2n+(2n-1)2n+1
则 -Tn=1×2+(2×22+2×23+···+2×2n)-(2n-1)2n+1,
即:-Tn=1×2+(23+24+····+2n+1)-(2n-1)2n+1,
∴Tn=(2n-3)2n+1+6
略
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